模板和std ::对列表初始化 [英] Templates and std::pair list-initialization

问题描述

这是此问题。

为什么编译：

#include <iostream>
class Test {
    public:
        Test(std::pair<char *, int>) {
            std::cout << "normal constructor 2!" << std::endl;
        }
};

int main() {
    Test t6({"Test", 42});
    return 0;
}

但这不会：

#include <iostream>
class Test {
    public:
        Test(std::pair<char *, int>) {
            std::cout << "normal constructor 2!" << std::endl;
        }
        template<typename ... Tn>
        Test(Tn ... args) {
            std::cout << "template constructor!" << std::endl;
        }
};

int main() {
    Test t6({"Test", 42});
    return 0;
}

错误讯息：

错误：调用'Test'的构造函数是不明确的

error: call to constructor of 'Test' is ambiguous

，如果它非常匹配，则非模板构造函数是优选的。所以我猜{Test，42}与std :: pair不匹配？
如果是这样，什么是正确的方式？我知道有std :: make_pair，但我想要尽可能短，因为我可以有几个这些对，并键入std :: make_pair每次都会不利，因为它膨胀的东西。

As I understood in the previous question, the non-template constructor is preferred, if it exactly matches. So I guess {"Test", 42} does not match with std::pair? If so, what would be the correct way? I know there is std::make_pair, but I want it as short as possible, because I can have several of those pairs and typing std::make_pair everytime would be unfavorable, because it bloats things. So what is the shortest way possible?

推荐答案

因为你已经在使用c ++ 11 ，切换到大括号初始化，你的代码将编译在gcc（至少4.7.2），并做你想要的：

Since you're already using c++11, switch to brace initialization and your code will compile under gcc (4.7.2 at least) and do what you want:

...
int main() {
    Test t6{{"Test", 42}};
    return 0;
}

$ g++ t.cpp -std=c++11
t.cpp: In function ‘int main()’:
t.cpp:14:25: warning: deprecated conversion from string constant to ‘char*’ [-Wwrite-strings]

$ a.out
normal constructor 2!

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